A more consistent definition of tetration of tetration for rational exponents
#3
Gottfried Wrote:Hi -

I'd like to try your computation with Pari/GP. Unfortunately I don't understand your maple-code completely. There are two calls of #-preceded expressions, #flip and #reduce, without parameters.

Gottfried

Hi Gottfried,

Trying the code is almost out of the question (because there is no code). I did all those by hand, by solving the corresponding equation with Maple. The Maple code which is displayed in the web page has nothing to do with it. This code just proves that any rational exponent m/n<1 is always a function of tetraroot exponents. One has to do the reduction to tetraroots by hand first and then solve the corresponding equation numerically.

Please look at the example. For the example, the equation to be solved with Maple is:

\( ((a^{(^{1/3}a)})^a)^{a^{(^{1/3}a)}}=e \)

Unfortunately, for every rational, the equation to be solved is different. For example, to calculate \( ^{9/10}e \), you would do:

\( ^{9/10}e=a \Leftrightarrow\\
e={^{10/9}}a \Leftrightarrow\\
e=a^{(^{1/9}a)} \)

so the equation to be solved numerically is the last one.
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Messages In This Thread
RE: A more consistent definition of tetration of tetration for rational exponents - by UVIR - 09/30/2007, 11:26 AM

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